Whole roots and powers

  • Thread starter georg gill
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Main Question or Discussion Point

[tex](a^{\frac{1}{2n}}a^{\frac{1}{2n}})^n=(a^{\frac{1}{2n}})^n(a^{\frac{1}{2n}})^n=(a^{\frac{1}{2}}) (a^{\frac{1}{2}})=a=(a^{\frac{1}{n}})^n[/tex]

all above is just done by using that the order of the factors that you multiply does not matter

we have proven that

[tex](a^{\frac{1}{p}}a^{\frac{1}{p}})^n=(a^{\frac{2}{p}})^n[/tex]

for any even number. Try saying that p is a odd number then n is a decimal number and taking the rooth with a number that is not whole does not make sense in a logixal approach of what one can really comprehend. For any even number of p what I did above makes sense. Could someone see an extention of this system. I want to make left side here:



[tex]\sqrt[m]{\underbrace{a^{\frac{1}{n}} \, \cdot \, a^{\frac{1}{n}} \, \cdot \, . . . \, \cdot a^{\frac{1}{n}} \, }_{\text{m times}} \, }=a^{\frac{m}{n}} [/tex]

to become right side by only using roths and powers of integers.
 
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Answers and Replies

  • #2
HallsofIvy
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Certainly [itex]a^{1/2n}a^{1/2n}= a^{1/2n+ 1/2n}= a^{2/2n}= a^{1/n}[/itex].

And [itex]\left(a^{1/n}\right)^m= a^{(1/n)m}= a^{m/n}[/itex]. Those are well known properties of the exponentials.
 
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  • #3
D H
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If you know what Dedekind cuts are, the PlanetMath page http://planetmath.org/encyclopedia/ProofOfPropertiesOfTheExponential.html [Broken] does a nice job of extending exponentiation from the integers to the rationals, and then to the reals.
 
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  • #4
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If you know what Dedekind cuts are, the PlanetMath page http://planetmath.org/encyclopedia/ProofOfPropertiesOfTheExponential.html [Broken] does a nice job of extending exponentiation from the integers to the rationals, and then to the reals.

As far as I can see or from what I get from the proof at least

[tex](xy)^m)^{\frac{1}{n}}=(y)^m)^{\frac{1}{n}}(x)^m)^{\frac{1}{n}}[/tex]

I want to prove that

[tex](xy)^m)^{\frac{1}{n}}=y)^{\frac{m}{n}}x^{\frac{m}{n}}[/tex]

What I have managed to show is that

[tex](xy^{\frac{1}{np}})^{np}[/tex]

[tex](a^{\frac{1}{np}})^{np}=a[/tex]

[tex]((a^{\frac{1}{n}})^{\frac{1}{p}})^{np}=(((a^{\frac{1}{n}})^{\frac{1}{p}})^n)^p[/tex]

since n and p are integers to put p outside as own power goues without proving (could of course prove this
and we get

[tex]((a^{\frac{1}{n}})^{\frac{1}{p}})^{np}=(((a^{\frac{1}{n}})^{\frac{1}{p}})^n)^p=a[/tex]

so

[tex]xy^{\frac{1}{np}}=(a^{\frac{1}{n}})^{\frac{1}{p}}[/tex]
 
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